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Physics Part-II

Nuclei

Obtain the binding energy of the nuclei ^{56}_{26}Fe and ^{209

Physics Part-II

Class 12

NCERT

Chapter 1: Ray Optics and Optical Instruments

41 questions

Chapter 2: Wave Optics

30 questions

Chapter 3: Dual Nature of Radiation and Matter

44 questions

Chapter 4: Atoms

23 questions

Chapter 5: Nuclei

38 questions

Obtain the binding energy (in MeV) of a nitrogen nucleus

(_{7}^{14} N)

[Given: m

(_{7}^{14} N)

=14.00307 u.]

Obtain the binding energy of the nuclei

_{26}^{56} F e

and

_{83}^{209} B i

in units of MeV from the following data:
(a)

m (_{26}^{56} F e) = 55.934939 u

(b)

m (_{83}^{209} B i) = 208.980388 u

A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin is entirely made of

_{29}^{63} C u

atoms (of mass 62.92960 u).

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Chapter 6: Semiconductor Electronics: Materials, Devices and Simple Circuits

25 questions

Question

Medium

Solving time: 4 mins

Obtain the binding energy of the nuclei $_{26}^{56} F e$ and $_{83}^{209} B i$ in units of MeV from the following data:
(a) $m (_{26}^{56} F e) = 55.934939 u$
(b) $m (_{83}^{209} B i) = 208.980388 u$

Name: Video solution 1: [Solved] Obtain the binding energy of the nuclei 2656Fe and 83209Bi in units of MeV from the following data:(a) m(2656Fe)=55.934939u(b) m(83209Bi)=208.980388u - Physics Part-II
Uploaded: 2023-04-23T05:55:01.000Z
Duration: 1 min
Description: Video solution 1: [Solved] Obtain the binding energy of the nuclei 2656Fe and 83209Bi in units of MeV from the following data:(a) m(2656Fe)=55.934939u(b) m(83209Bi)=208.980388u - Physics Part-II

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Updated on: Apr 23, 2023

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Physics Part-II (NCERT)

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Obtain the binding energy of the nuclei

_{26}^{56} F e

and

_{83}^{209} B i

in units of MeV from the following data:
(a)

m (_{26}^{56} F e) = 55.934939 u

(b)

m (_{83}^{209} B i) = 208.980388 u

15 mins ago

Discuss this question LIVE

15 mins ago

Text solutionVerified

(a)

_{26}^{56} F e

contains 26 protons and 30 neutrons.

Mass of 26 protons=

26 \times 1.007825 u = 26.26345 u

Mass of 30 neutrons=

30 \times 1.008665 u = 30.25995 u

Mass of

_{26}^{56} F e

nucleus=

55.934939 u

Thus, mass defect

Δ m = 26.26345 u + 30.25995 u - 55.934939 u = 0.528461 u

Thus, binding energy of

_{26}^{56} F e

nucleus is

0.528461 \times 931.5 M e V = 492.26 M e V

Binding energy per nucleon=

\frac{492.26 M e V}{56} = 8.790 M e V

(b)

_{83}^{209} B i

contains 83 protons and 126 neutrons.

Mass of 83 protons=

83 \times 1.007825 u = 83.649475 u

Mass of 126 neutrons=

126 \times 1.008665 u = 127.09170 u

Mass of

_{83}^{209} B i

nucleus=

208.986388 u

Thus, mass defect

Δ m = 83.649475 u + 127.09170 u - 208.986388 u = 1.760872 u

Thu,s binding energy of

_{83}^{209} B i

nucleus is

1.760872 \times 931.5 M e V = 1640.26 M e V

Binding energy per nucleon=

\frac{1640.26 M e V}{209} = 7.848 M e V

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Practice more questions from Physics Part-II (NCERT)

Obtain the binding energy (in MeV) of a nitrogen nucleus

(_{7}^{14} N)

[Given: m

(_{7}^{14} N)

=14.00307 u.]

Obtain the binding energy of the nuclei

_{26}^{56} F e

and

_{83}^{209} B i

in units of MeV from the following data:
(a)

m (_{26}^{56} F e) = 55.934939 u

(b)

m (_{83}^{209} B i) = 208.980388 u

A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin is entirely made of

_{29}^{63} C u

atoms (of mass 62.92960 u).

View all

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Question Text	Obtain the binding energy of the nuclei $_{26}^{56} F e$ and $_{83}^{209} B i$ in units of MeV from the following data: (a) $m (_{26}^{56} F e) = 55.934939 u$ (b) $m (_{83}^{209} B i) = 208.980388 u$
Updated On	Apr 23, 2023
Topic	Nuclei
Subject	Physics
Class	Class 12
Answer Type	Text solution:1 Video solution: 1
Upvotes	187
Avg. Video Duration	1 min

Obtain the binding energy of the nuclei $_{26}^{56} F e$ and $_{83}^{209} B i$ in units of MeV from the following data:
(a) $m (_{26}^{56} F e) = 55.934939 u$
(b) $m (_{83}^{209} B i) = 208.980388 u$

Text solutionVerified

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Obtain the binding energy of the nuclei $_{26}^{56} F e$ and $_{83}^{209} B i$ in units of MeV from the following data:
(a) $m (_{26}^{56} F e) = 55.934939 u$
(b) $m (_{83}^{209} B i) = 208.980388 u$